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Difference between revisions of "Talk:2021 Fall AMC 12B Problems/Problem 17"

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==Solution 3==
 
==Solution 3==
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<math></math>
 
 
Use generating function, define <math>c_{n}\cdot x^{n}</math> be <math>c_{n}</math> ways for the end point be <math>{n}</math> unit away from the origins.  
 
Use generating function, define <math>c_{n}\cdot x^{n}</math> be <math>c_{n}</math> ways for the end point be <math>{n}</math> unit away from the origins.  
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<math></math>
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Therefore,
 
Therefore,
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if the current point is origin, need to <math>\cdot6{x}</math>
 
if the current point is origin, need to <math>\cdot6{x}</math>
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\\if the current point on vertex of the unit hexagon, need to <math>\cdot(x^{-1}+2)</math>, where there is one way to return to the origin and there are two ways to keep distance = 1
 
  
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\\Now let's start
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if the current point on vertex of the unit hexagon, need to <math>\cdot(x^{-1}+2)</math>, where there is one way to return to the origin and there are two ways to keep distance = 1
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\\init <math>p(x)=1</math>;   
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\\1st step:  <math>p(x)=6x</math>\r\n    
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Now let's start with <math>p(x)=1</math>;   
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\\2nd step:  <math>p(x)=6x\cdot(x^{-1}+2) = 6 + 12x </math>
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\\3rd step:  <math>p(x)=6\cdot6x + 12x\cdot(x^{-1}+2) = 12 + 60x</math>
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1st step:  <math>p(x)=6x</math>   
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\\4th step:  <math>p(x)=12\cdot6x + 60x\cdot(x^{-1}+2) = 60 + 192x </math>
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\\5th step:  <math>p(x)=60\cdot6x + 192x\cdot(x^{-1}+2) = 192 + 744x </math>
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2nd step:  <math>p(x)=6x\cdot(x^{-1}+2) = 6 + 12x </math>
 +
 
 +
3rd step:  <math>p(x)=6\cdot6x + 12x\cdot(x^{-1}+2) = 12 + 60x</math>
 +
 
 +
4th step:  <math>p(x)=12\cdot6x + 60x\cdot(x^{-1}+2) = 60 + 192x </math>
 +
 
 +
5th step:  <math>p(x)=60\cdot6x + 192x\cdot(x^{-1}+2) = 192 + 744x </math>
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So there are <math>192+744=936</math> ways for the bug never moves more than 1 unit away from orign, and <math>\frac{936}{6^5} = \boxed{\frac{13}{108}}.</math>
  
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\\So there are <math>192+744=936</math> ways for the bug never moves more than 1 unit away from orign, and <math>\frac{936}{6^5} = \boxed{\frac{13}{108}}.</math>
 
  
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%
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-wwei.yu
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\\-wwei.yu
 
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\end{left}
 

Latest revision as of 17:52, 28 November 2021

Solution 3

Use generating function, define $c_{n}\cdot x^{n}$ be $c_{n}$ ways for the end point be ${n}$ unit away from the origins.

Therefore,

if the current point is origin, need to $\cdot6{x}$

if the current point on vertex of the unit hexagon, need to $\cdot(x^{-1}+2)$, where there is one way to return to the origin and there are two ways to keep distance = 1

Now let's start with $p(x)=1$;

1st step: $p(x)=6x$

2nd step: $p(x)=6x\cdot(x^{-1}+2) = 6 + 12x$

3rd step: $p(x)=6\cdot6x + 12x\cdot(x^{-1}+2) = 12 + 60x$

4th step: $p(x)=12\cdot6x + 60x\cdot(x^{-1}+2) = 60 + 192x$

5th step: $p(x)=60\cdot6x + 192x\cdot(x^{-1}+2) = 192 + 744x$

So there are $192+744=936$ ways for the bug never moves more than 1 unit away from orign, and $\frac{936}{6^5} = \boxed{\frac{13}{108}}.$


-wwei.yu

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